TEST OF PRACTICAL KNOWLEDGE QUESTION You are provided with a potentiometer AB, a \(102\Omega\) standard resistor R, a battery of emf 4.5V, a jockey J, and o...

Answer Details

Test of Practical Knowledge: Potentiometer (Electricity)

The circuit is connected as shown below. The 4.5 V battery drives a current through the key K, the ammeter, the standard resistor R (102 Ω) and the length AJ = x of the potentiometer wire. As the jockey J is slid away from A the length x (and therefore the resistance in the circuit) increases, so the ammeter reading I falls.

With the key closed but the jockey not touching AB, the ammeter reads

\[ I = 0.36\,\text{A}. \]

Table of readings

For each position of the jockey the length x is read, the key is closed and the ammeter reading I is recorded, then \(x^{-1}=1/x\) is evaluated:

x / cm	I / A	\(x^{-1}\) / cm\(^{-1}\)
20.0	1.14	0.050
35.0	0.90	0.029
45.0	0.84	0.022
60.0	0.78	0.017
80.0	0.66	0.013

Graph of \(x^{-1}\) against I

\(x^{-1}\) is plotted on the vertical axis and I on the horizontal axis, both axes starting from the origin (0, 0):

Slope of the graph

Taking two widely separated points that lie on the line of best fit, \((I_1,\,x_1^{-1})=(0.70,\,0.013)\) and \((I_2,\,x_2^{-1})=(1.10,\,0.045)\):

\[ s=\frac{\Delta(x^{-1})}{\Delta I}=\frac{0.045-0.013}{1.10-0.70}=\frac{0.032}{0.40}=0.080\ \text{cm}^{-1}\,\text{A}^{-1}. \]

Value \(I_0\) of I for which \(x^{-1}=0\)

Producing the line of best fit until it cuts the horizontal (I) axis, the point where \(x^{-1}=0\) is

\[ I_0=0.54\,\text{A}. \]

Evaluate \(\dfrac{I_0}{I}\)

\[ \frac{I_0}{I}=\frac{0.54}{0.36}=1.5. \]

Two precautions

1. The key was opened immediately after each reading was taken, so that current did not flow continuously and heat the wire or run down the battery.
2. Errors of parallax were avoided by viewing the metre scale and the ammeter pointer with the line of sight perpendicular to the scale, and all connections were kept clean and tight.

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(b)(i) Definition of e.m.f.

The e.m.f. of a battery is the total energy supplied by the battery per unit charge in driving that charge round the complete circuit. Equivalently, it is the potential difference across the terminals of the battery when it is on open circuit, i.e. when it is not supplying current to an external circuit.

(b)(ii) e.m.f. of cell Y

When two cells are balanced in turn against the same potentiometer wire carrying the same steady current, the e.m.f. of a cell is directly proportional to its balance length, \(E\propto L\). Hence

\[ \frac{E_X}{E_Y}=\frac{L_X}{L_Y}. \]

With \(E_X=1.018\,\text{V}\), \(L_X=50.0\,\text{cm}\) and \(L_Y=75.0\,\text{cm}\):

\[ E_Y=E_X\times\frac{L_Y}{L_X}=1.018\times\frac{75.0}{50.0}=1.53\,\text{V}. \]

Since cell Y balances at the greater length it has the greater e.m.f.:

\[ E_Y=1.53\,\text{V}. \]